Stopping Strategies and Gambler's Ruin
| Autor: | James D. Harper, Kenneth A. Ross |
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| Rok vydání: | 2005 |
| Předmět: | |
| Zdroj: | Mathematics Magazine. 78:255-268 |
| ISSN: | 1930-0980 0025-570X |
| DOI: | 10.1080/0025570x.2005.11953355 |
| Popis: | Let's play a game. Roll a die and you win $2 if the die shows 1 or 2. Otherwise you lose $1. Thus about one-third of the time you win $2 and about two-thirds of the time you lose $1. Suppose you have a definite idea in mind about how much you would like to win. To be concrete, suppose you start with $5 and decide to play until you double your fortune or lose it all (that is, are ruined). In other words, you will stop playing when you have $0, $10, or $11. This game with a die is fair because, on average, for each 3 games, you win $2 once and lose $1 twice. However, it is not the same fair game as in the classical coinflipping game where you win or lose $1, each with probability 1/2. One difference is the experience of a player facing ruin at the gambling table. The classical theory of what is called "gambler's ruin" tells us that the probability of ruin in the coin-flipping game is 1/2, that is, a player who starts with $5 and swears to stop at $10-a doubleor-nothing strategy-faces even chances of being ruined or winning double. Knowing this about the coin-flipping game calls some parallel questions to mind. In our die-rolling game, is it still true that probabilities for success or ruin are 1/2, that the double-or-nothing strategy is fair? What is the expected time for the duration of this strategy? In other words, how many games will you play on average? We ask similar questions for unfair games. For example, we consider the game where you win $2 with probability 4/40, you win $1 with probability 11/40, and you lose $1 with probability 25/40. This game is not fair; the average loss per game is 15 cents. |
| Databáze: | OpenAIRE |
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